二维带形区域上Chemotaxis-Navier-Stokes方程的整体适定性

doi:10.12677/pm.2024.144129

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二维带形区域上Chemotaxis-Navier-Stokes方程的整体适定性
Global Well-Posedness for the Chemotaxis-Navier-Stokes Equations on a 2-D Strip Domain

DOI: 10.12677/pm.2024.144129, PDF, HTML, XML, 下载: 98 浏览: 244
作者: 郭晶, 刘晓风：东华大学理学院，上海
关键词: Chemotaxis-Navier-Stokes方程；带形区域；先验估计；解的整体存在性；Chemotaxis-Navier-Stokes Equation； A Strip Domain； Prior Estimate； The Global Existence of the Solution

摘要: 本文研究了二维带形区域上带齐次Neumann-Neumann-Dirichlet边界条件的Chemotaxis-Navier-Stokes方程解的适定性问题。当该方程在平衡态

附近满足一定的初始条件和假设条件

时，通过建立能量泛函和利用一些不等式的方法得到该方程解的一致先验估计；最后再结合局部存在性和连续性证明了解的整体存在性。

Abstract: In this paper, the solution of the Chemotaxis-Navier-Stokes equation with homogeneous Neumann-Neumann-Dirichlet boundary conditions over a two-dimensional strip domain is studied. When the equation satisfies certain initial conditions and assumptions

near the equilibrium state

, the uniform prior estimates of the solution of the equation are obtained by establishing energy functional and using some inequalities. Finally, the global existence of solution is proved by the combination of local existence and continuity.

文章引用：郭晶, 刘晓风. 二维带形区域上Chemotaxis-Navier-Stokes方程的整体适定性[J]. 理论数学, 2024, 14(4): 213-228. https://doi.org/10.12677/pm.2024.144129

1. 引言

趋化–流体耦合模型起源于Tuval等人 [1] 对枯草芽孢杆菌这样的趋氧型细菌在水滴中集中悬浮的实验观察，他们首次在 $Ω \times ℝ^{+}$ 上提出如下用来描述氧气驱动的枯草芽孢杆菌在不可压缩流体中游动的趋化–流体耦合模型

${\begin{cases} n_{t} + u \cdot \nabla n = D_{n} Δ n - \nabla \cdot (n χ (c) \nabla c), \\ c_{t} + u \cdot \nabla c = D_{c} Δ c - n f (c), \\ u_{t} + κ (u \cdot \nabla u) + \nabla P = D_{u} Δ u - n \nabla φ, \\ \nabla \cdot u = 0, \end{cases}$ (1.1)

这里n、c和 $u$ 分别表示细菌密度、氧气的浓度和流体速度，标量函数 $P = P (x, t)$ 表示压力， $χ (c)$ 表示氧气浓度变化引起趋化的敏感度函数， $φ$ 是重力势函数，正常数 $D_{n}, D_{c}$ 和 $D_{u}$ 分别表示细菌、氧气和流体的扩散系数，而参数 $κ$ 是非负常数且与非线性对流强度有关，通常用 $κ = 0$ 表示Stokes流， $κ = 1$ 表示Navier-Stokes流。

现回顾模型(1.1)相关的一些研究成果。文献 [2] 在二维全空间中研究了四元Chemotaxis-Navier-Stokes方程的全局经典解，文献 [3] 在具有单向缓变的大初值条件下研究了三维全空间中该模型解的整体适定性。对于 $Ω$ 为有界区域的情形，当 $Ω$ 的边界光滑时，需要对边界施加一些适当的限制，最常见的一种边界条件是 $n, c$ 和 $u$ 分别满足齐次Neumann-Neumann-Dirichlet边界条件，即

$\frac{\partial n}{\partial ν} = \frac{\partial c}{\partial ν} = 0, u = 0, x \in \partial Ω,$ (1.2)

其中 $ν$ 为区域边界处的单位外法向量。在这种条件下，文献 [4] [5] [6] 在对f和 $χ$ 进行一定的结构性假设的情况下证明了模型(1.1)的适定性结果及解的相关性质。文献 [7] 研究了小初值条件下趋化Navier-Stokes方程的全局经典解。随后，文献 [8] 在没有小初值的条件下研究了二维有界区域上模型(1.1)解的整体存在性和稳定性问题。然而目前关于趋化–流体耦合模型在带状区域上解的适定性及相关性质的研究还很有限。文献 [9] [10] 在二维矩形区域和三维平行六面体区域的情形下对Chemotaxis-Navier-Stokes系统进行了数值模拟实验。文献 [11] [12] 在三维无界区域中得到了此模型强解的整体存在性理论。但是在二维带形区域上还没有类似的研究结果。因此，与文献 [11] 不同，本文改进了小初值的条件，在二维带形区域上研究了模型(1.1)解的整体适定性问题。

本文将在 $Ω \times ℝ^{+}$ 上考虑如下Chemotaxis-Navier-Stokes系统：

${\begin{cases} n_{t} + u \cdot \nabla n = Δ n - \nabla \cdot (n χ (c) \nabla c), \\ c_{t} + u \cdot \nabla c = Δ c - n f (c), \\ u_{t} + u \cdot \nabla u + \nabla P = Δ u - n \nabla φ, \\ \nabla \cdot u = 0, \end{cases}$ (1.3)

其中 $Ω = {x = (x_{1}, x_{2}) \in ℝ^{2} | x_{1} \in ℝ, 0 \leq x_{2} \leq 1}$ ，在区域 $Ω$ 的上边界 $Γ_{B} = {(x_{1}, x_{2}) \in ℝ^{2} | x_{1} \in ℝ, x_{2} = 0}$ 和下边界 $Γ_{T} = {(x_{1}, x_{2}) \in ℝ^{2} | x_{1} \in ℝ, x_{2} = 1}$ 上考虑通常的边界条件(1.2)，满足如下假设条件：

${\begin{cases} 初始条件 {(n, c, u)}_{t = 0} = (n_{0} (x), c_{0} (x), u_{0} (x)) 满足 \\ n_{0} (x) \geq 0, c_{0} (x) \geq 0, \nabla \cdot u_{0} (x) = 0 对任意的 x \in Ω 成立, \\ χ, f 和 φ 都是光滑函数且 f (0) = 0, 对 \forall s \in ℝ 有 f^{'} (s) \geq 0 成立, \\ φ \in W^{1, \infty} (Ω) 且 {‖ c_{0} ‖}_{L^{\infty} (Ω)} 充分小 . \end{cases}$ (1.4)

本文目标是要证明当初值 $(n_{0} (x), c_{0} (x), u_{0} (x))$ 在稳态 $(n_{\infty}, 0, 0) (n_{\infty} \geq 0)$ 附近作光滑小扰动时方程组(1.3)解的整体存在性。为了简便，取变量变换，令 $n = λ + n_{\infty}$ 和 $\tilde{p} = p + n_{\infty} φ$ ，系统(1.3)可改写为

${\begin{cases} λ_{t} + u \cdot \nabla λ = Δ λ - n_{\infty} \nabla \cdot (χ (c) \nabla c) - \nabla \cdot (χ (c) λ \nabla c), \\ c_{t} + u \cdot \nabla c = Δ c - (λ + n_{\infty}) f (c), \\ u_{t} + u \cdot \nabla u + \nabla \tilde{P} = Δ u - λ \nabla φ, \\ \nabla \cdot u = 0, \end{cases}$ (1.5)

由于在 $Γ_{T}$ 上 $ν = (0, 1)$ ，而在 $Γ_{B}$ 上 $ν = (0, - 1)$ ，故初边值条件满足

${\begin{cases} {(λ, c, u)}_{t = 0} = (λ_{0} (x), c_{0} (x), u_{0} (x)), x \in Ω \\ \partial_{2} λ = \partial_{2} c = 0, u = 0, x \in Γ_{B} \cup Γ_{T}, t > 0 \end{cases}$ (1.6)

其中 $λ_{0} = n_{0} - n_{\infty}$ 。本文的主要结果如下所述：

定理1.1 假设条件(1.4)成立，并且存在一个充分小的常数 $ε_{0} > 0$ 使得当 ${‖ (λ_{0}, c_{0}, u_{0}) ‖}_{H^{1} (Ω)} + {‖ \nabla (λ_{0}, c_{0}, u_{0}) ‖}_{L^{4} (Ω)} \leq ε_{0}$ 时，方程组(1.5)~(1.6)有唯一强解 $(λ, c, u)$ 满足 $n (x, t) \equiv λ (x, t) + n_{\infty} \geq 0$ ， $c (x, t) \geq 0$ ， $x \in Ω$ ， $t \geq 0$ ，并且存在一个正常数 $C_{0}$ 使得

$\begin{array}{l} ({‖ (λ, c, u) (\cdot, t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ, c, u) ‖}_{L^{4} (Ω)}^{4}) \\ + \int_{0}^{t} ({‖ (\nabla λ, \nabla c, \nabla u) (\cdot, s) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ, c, u) \cdot Δ (λ, c, u) ‖}_{L^{2} (Ω)}^{2}) d s \\ \leq C_{0} ({‖ (λ_{0}, c_{0}, u_{0}) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ_{0}, c_{0}, u_{0}) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$ (1.7)

对任意的 $t \geq 0$ 成立。

2. 预备知识

在开始证明之前，我们首先给出本文需要用到的一些重要引理 [13] [14] 。

引理2.1 (Poincaré不等式)设 $1 \leq p < + \infty$ 且 $Ω \subset ℝ^{N}$ 为有界区域，那么

1) 如果 $u \in W_{0}^{1, p} (Ω)$ ，则有 $\int_{Ω} {| u |}^{p} d x \leq C \int_{Ω} {| D u |}^{p} d x$ ；

2) 如果 $\partial Ω$ 满足局部Lipschitz条件， $u \in W^{1, p} (Ω)$ ，则有 $\int_{Ω} {| u - u_{Ω} |}^{p} d x \leq C \int_{Ω} {| D u |}^{p} d x$ ，其中C是依赖于 $n, p, Ω$ 的常数， $u_{Ω} = \frac{1}{| Ω |} \int_{Ω} u (x) d x$ ，这里的 $| Ω |$ 表示 $Ω$ 的测度。

注：如果 $Ω \subset ℝ^{2}$ 为带形区域，也有类似的结果，具体可参见文献 [14] 。

引理2.2 假设条件(1.4)成立，那么初边值问题(1.3)的解 $(n, c, u)$ 满足 $n (x, t) \geq 0$ ， $c (x, t) \geq 0$ ， $x \in Ω$ ， $t > 0$ ，和 ${‖ n (\cdot, t) ‖}_{L^{1} (Ω)} \equiv {‖ n_{0} ‖}_{L^{1} (Ω)}$ ， $t \geq 0$ ，并且对任意的 $1 \leq p \leq + \infty$ ，有 $\sup_{t \geq 0} {‖ c (\cdot, t) ‖}_{L^{p} (Ω)} \leq {‖ c_{0} ‖}_{L^{p} (Ω)}$ ， $t \geq 0$ 。

证明：详细证明过程可参见文献 [11] 。

注：在本文中， $c_{\infty} = {‖ c_{0} ‖}_{L^{\infty} (Ω)}$ ， $M_{f} = \max_{0 \leq s \leq c_{\infty}} | f^{'} (s) | + \max_{0 \leq s \leq c_{\infty}} | f^{″} (s) |$ ， $M_{χ} = \max_{0 \leq s \leq c_{\infty}} | χ (s) | + \max_{0 \leq s \leq c_{\infty}} | χ^{'} (s) |$ ；除特别说明外， $C_{i} (i = 1, 2, \dots)$ 均表示与未知函数和时间无关的一般正常数。

3. 局部存在唯一性

为了证明方程组(1.5)~(1.6)强解的局部存在唯一性，我们首先对其构造一个线性迭代格式如下

${\begin{cases} λ_{t}^{j + 1} - Δ λ^{j + 1} = - u^{j} \cdot \nabla λ^{j + 1} - n_{\infty} \nabla \cdot (χ (c^{j}) \nabla c^{j + 1}) - \nabla \cdot (χ (c^{j}) λ^{j} \nabla c^{j + 1}), \\ c_{t}^{j + 1} - Δ c^{j + 1} = - u^{j} \cdot \nabla c^{j + 1} - (λ^{j + 1} + n_{\infty}) f (c^{j}), \\ u_{t}^{j + 1} - Δ u^{j + 1} + \nabla {\tilde{P}}^{j + 1} = - u^{j} \cdot \nabla u^{j + 1} - λ^{j} \nabla φ, \\ \nabla \cdot u^{j + 1} = 0, \end{cases}$ (3.1)

以及初边值条件

${\begin{cases} {(λ^{j + 1}, c^{j + 1}, u^{j + 1})}_{t = 0} = (λ_{0} (x), c_{0} (x), u_{0} (x)), x \in Ω \\ \partial_{2} λ^{j + 1} = \partial_{2} c^{j + 1} = 0, u^{j + 1} = 0, x \in Γ_{B} \cup Γ_{T}, t > 0 \end{cases}$ (3.2)

其中 $j \geq 0$ ， $(λ^{0}, c^{0}, u^{0}) : = (0, 0, 0)$ 。为了方便，我们记 $S^{j} = (λ^{j}, c^{j}, u^{j})$ ， $S (0) : = (λ_{0}, c_{0}, u_{0})$ ， $j \geq 0$ .从而，有以下局部存在性定理。

引理3.1 假设定理1.1中的所有条件成立，设 ${‖ S (0) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq ε_{1}$ ，则存在正常数 $T_{0}, M_{0}$ 使得对于任意的 $j \geq 0$ ， $S (0) \in C ([0, T_{0}]; H^{1} (Ω) \cap W^{1, 4} (Ω))$ 都是适定的，并且满足一致估计

$\sup_{0 \leq t \leq T_{0}} {‖ S^{j} (t) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq M_{0}$ ， $j \geq 0$ (3.3)

此外， ${S^{j}}_{j \geq 0}$ 是Banach空间 $C ([0, T_{0}]; H^{1} (Ω) \cap W^{1, 4} (Ω))$ 中的一个Cauchy序列，其对应的极限 $S : = (λ, c, u) \in C ([0, T_{0}]; H^{1} (Ω) \cap W^{1, 4} (Ω))$ 满足

$\sup_{0 \leq t \leq T_{0}} {‖ S (t) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq M_{0}$ ， (3.4)

并且 $S = (λ, c, u)$ 是系统(1.5)~(1.6)在 $[0, T_{0}]$ 上的唯一解。

证明：我们将引理3.1的证明分为两步。

第一步： ${S^{j}}_{j \geq 0}$ 在 $[0, T_{0}]$ 上的一致有界估计。我们将运用数学归纳法来证明。当 $j = 0$ 时，根据初始假设条件 $S^{0} = (λ^{0}, c^{0}, u^{0})$ 可知不等式(3.3)成立。假设对于每一个 $j \geq 0$ ，不等式(3.3)成立，其中 $M_{0} \in (0, 1)$ 充分小，然后证明当 $j + 1$ 时不等式(3.3)仍然成立。为此，我们需要得到 $S^{j + 1}$ 的一些能量估计。

1) ( $λ^{j + 1}$ , $c^{j + 1}$ , $u^{j + 1}$ )的 $L^{2}$ 估计

在方程(3.1)₁的左右两端同时乘以 $λ^{j + 1}$ 并在 $Ω$ 上分部积分得

$\frac{1}{2} \frac{d}{d t} {‖ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} = \int_{Ω} (n_{\infty} + λ^{j}) χ (c^{j}) \nabla c^{j + 1} \cdot \nabla λ^{j + 1} d x,$

根据条件(3.2)，运用Hölder不等式、Young不等式、Gagliardo-Nirenberg插值不等式和Sobolev嵌入定理得

$\begin{array}{l} \int_{Ω} (n_{\infty} + λ^{j}) χ (c^{j}) \nabla c^{j + 1} \cdot \nabla λ^{j + 1} d x \\ \leq n_{\infty} M_{χ} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)} + M_{χ} {‖ λ^{j} ‖}_{L^{4} (Ω)} {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq C ({‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)} + {‖ λ^{j} ‖}_{H^{1} (Ω)} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}) \\ \leq \frac{M_{0}}{2} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + \frac{C}{2} ({‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2}) . \end{array}$

从而，有

$\frac{1}{2} \frac{d}{d t} {‖ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} \leq \frac{M_{0}}{2} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + \frac{C}{2} ({‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2})$ (3.5)

同样地，运用假设条件 $f (0) = 0$ ，我们可以得到 $c^{j + 1}$ ， $u^{j + 1}$ 的 $L^{2}$ 模估计

$\frac{1}{2} \frac{d}{d t} {‖ c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} \leq C ({‖ c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ λ^{j + 1} ‖}_{H^{1} (Ω)}^{2} + 1) .$ (3.6)

$\frac{1}{2} \frac{d}{d t} {‖ u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u^{j + 1} ‖}_{L^{2} (Ω)}^{2} \leq C M_{0} ({‖ u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + 1) .$ (3.7)

2) ( $λ^{j + 1}$ , $c^{j + 1}$ , $u^{j + 1}$ )的 $H^{1}$ 估计

对于 $λ^{j + 1}$ 的 $H^{1}$ 模估计，我们取方程(3.1)₁与 $- Δ λ^{j + 1}$ 的 $L^{2}$ 内积，分部积分得

$\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} \\ = \int_{Ω} (u^{j} \cdot \nabla λ^{j + 1}) \cdot Δ λ^{j + 1} d x + n_{\infty} \int_{Ω} \nabla (χ (c^{j}) \nabla c^{j + 1}) \cdot Δ λ^{j + 1} d x + \int_{Ω} \nabla \cdot (χ (c^{j}) λ^{j} \nabla c^{j + 1}) \cdot Δ λ^{j + 1} d x . \end{array}$ (3.8)

再次利用Hölder不等式、Young不等式、Gagliardo-Nirenberg插值不等式和Poincaré不等式，可得到(3.8)式右端的积分项估计如下

对于第一项有

$\begin{matrix} \int_{Ω} (u^{j} \cdot \nabla λ^{j + 1}) \cdot Δ λ^{j + 1} d x \leq {‖ u^{j} ‖}_{L^{4} (Ω)} {‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq C {‖ u^{j} ‖}_{H^{1} (Ω)} {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq \frac{C M_{0}}{2} {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{} + \frac{M_{0}}{2} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} \\ \leq M_{0} {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + \frac{C}{4} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} . \end{matrix}$ (3.9)

对于第二项有

$\begin{array}{l} n_{\infty} \int_{Ω} \nabla (χ (c^{j}) \nabla c^{j + 1}) \cdot Δ λ^{j + 1} d x \\ \leq n_{\infty} M_{χ} {‖ \nabla c^{j} ‖}_{L^{4} (Ω)} {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} + n_{\infty} M_{χ} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq C (M_{0} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} + {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}) \\ \leq M_{0} ({‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + \frac{C}{2} ({‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + 1) . \end{array}$ (3.10)

对于第三项有

$\begin{array}{l} \int_{Ω} \nabla \cdot (χ (c^{j}) λ^{j} \nabla c^{j + 1}) \cdot Δ λ^{j + 1} d x \\ = \int_{Ω} χ^{'} (c^{j}) \nabla c^{j} λ^{j} \nabla c^{j + 1} \cdot Δ λ^{j + 1} d x + \int_{Ω} χ (c^{j}) \nabla λ^{j} \nabla c^{j + 1} \cdot Δ λ^{j + 1} d x \\ + \int_{Ω} χ (c^{j}) λ^{j} \nabla^{2} c^{j + 1} \cdot Δ λ^{j + 1} d x : = I_{1} + I_{2} + I_{3} . \end{array}$ (3.11)

其中

$\begin{matrix} I_{1} = M_{χ} {‖ \nabla c^{j} ‖}_{L^{4} (Ω)} {‖ λ^{j} ‖}_{L^{8} (Ω)} {‖ \nabla c^{j + 1} ‖}_{L^{8} (Ω)} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq C M_{0} {‖ λ^{j} ‖}_{H^{1} (Ω)} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{3}{4}} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{4}} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq \frac{M_{0}}{2} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + \frac{C M_{0}}{2} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{3}{2}} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} \\ \leq M_{0} ({‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + \frac{C}{4} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} . \end{matrix}$ (3.12)

$\begin{array}{l} I_{2} = M_{χ} {‖ \nabla λ^{j} ‖}_{L^{4} (Ω)} {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \leq C M_{0} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq \frac{M_{0}}{2} ({‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + \frac{C}{4} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (3.13)

和

$\begin{array}{l} I_{3} = M_{χ} {‖ λ^{j} ‖}_{L^{4} (Ω)} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{4} (Ω)} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \leq C M_{0} {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)} \\ \leq \frac{M_{0}}{2} ({‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + \frac{C M_{0}}{4} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (3.14)

将(3.12)式、(3.13)式和(3.14)代入(3.11)式，从而有

$\begin{array}{l} \int_{Ω} \nabla \cdot (χ (c^{j}) λ^{j} \nabla c^{j + 1}) \cdot Δ λ^{j + 1} d x \\ \leq 2 M_{0} ({‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + \frac{C M_{0}}{4} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + \frac{C}{2} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (3.15)

将(3.9)式、(3.10)式和(3.15)式代入(3.8)式，整理系数，对适当小的 $M_{0}$ 可得

$\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} \\ \leq 4 M_{0} ({‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + \frac{C M_{0}}{2} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C ({‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + 1) . \end{array}$ (3.16)

类似地，得到 $c^{j + 1}$ ， $u^{j + 1}$ 的 $H^{1}$ 模估计分别为

$\frac{1}{2} \frac{d}{d t} {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ c^{j + 1} ‖}_{L^{2} (Ω)}^{2} \leq M_{0} {‖ Δ c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C ({‖ λ^{j + 1} ‖}_{H^{1} (Ω)}^{2} + 1) .$ (3.17)

$\frac{1}{2} \frac{d}{d t} {‖ \nabla u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ u^{j + 1} ‖}_{L^{2} (Ω)}^{2} \leq 3 M_{0} {‖ Δ u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C ({‖ \nabla u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + 1) .$ (3.18)

3) ( $\nabla λ^{j + 1}$ , $\nabla c^{j + 1}$ , $\nabla u^{j + 1}$ )的 $L^{4}$ 估计

对于 $\nabla λ^{j + 1}$ 的 $L^{4}$ 模估计，我们先对方程(3.1)₁作用 $\nabla$ ，再将得到的方程两端同乘以 $4 {| \nabla λ^{j + 1} |}^{2} \nabla λ^{j + 1}$ ，然后在在 $Ω$ 上进行积分得

$\begin{array}{l} \frac{d}{d t} \int_{Ω} {| \nabla λ^{j + 1} |}^{4} d x + 4 \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + 2 \int_{Ω} {| \nabla ({| \nabla λ^{j + 1} |}^{2}) |}^{2} d x \\ = - 4 \int_{Ω} \nabla \cdot (u^{j} \cdot \nabla λ^{j + 1} + n_{\infty} \nabla \cdot (χ (c^{j}) \nabla c^{j + 1}) + \nabla \cdot (χ (c^{j}) λ^{j} \nabla c^{j + 1})) \cdot {| \nabla λ^{j + 1} |}^{2} \nabla λ^{j + 1} d x \\ = 4 \int_{Ω} u^{j} \cdot \nabla λ^{j + 1} \cdot ({| \nabla λ^{j + 1} |}^{2} \nabla^{2} λ^{j + 1} + \nabla ({| \nabla λ^{j + 1} |}^{2}) \cdot \nabla λ^{j + 1}) d x \\ + 4 \int_{Ω} n_{\infty} \nabla \cdot (χ (c^{j}) \nabla c^{j + 1}) \cdot ({| \nabla λ^{j + 1} |}^{2} \nabla^{2} λ^{j + 1} + \nabla ({| \nabla λ^{j + 1} |}^{2}) \cdot \nabla λ^{j + 1}) d x \\ + 4 \int_{Ω} \nabla \cdot (χ (c^{j}) λ^{j} \nabla c^{j + 1}) \cdot ({| \nabla λ^{j + 1} |}^{2} \nabla^{2} λ^{j + 1} + \nabla ({| \nabla λ^{j + 1} |}^{2}) \cdot \nabla λ^{j + 1}) d x \\ : = J_{1} + J_{2} + J_{3} . \end{array}$

对上式利用Hölder不等式、Gagliardo-Nirenberg插值不等式和Sobolev嵌入定理，得到

$\begin{array}{l} J_{1} \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + C \int_{Ω} {| u^{j} |}^{2} \cdot {| \nabla λ^{j + 1} |}^{4} d x \\ \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + C {‖ u^{j} ‖}_{L^{4} (Ω)}^{2} {‖ {| \nabla λ^{j + 1} |}^{2} ‖}_{L^{4} (Ω)}^{2} \\ \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + M_{0} {‖ {| \nabla λ^{j + 1} |}^{2} ‖}_{L^{4} (Ω)}^{4} + C M_{0} . \end{array}$

其中

$\begin{matrix} {‖ {| \nabla λ^{j + 1} |}^{2} ‖}_{L^{4} (Ω)}^{4} \leq C {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{} {‖ {| \nabla λ^{j + 1} |}^{2} ‖}_{L^{3} (Ω)}^{3} \\ \leq {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + \frac{C}{2} {‖ \nabla λ^{j + 1} ‖}_{L^{6} (Ω)}^{3} \\ \leq {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + \frac{C}{2} {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{} {‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{2} \\ \leq {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C {‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4} \end{matrix}$

因此， $J_{1} \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + M_{0} {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C M_{0} ({‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4} + 1)$ .

$\begin{array}{l} J_{2} \leq C \int_{Ω} \nabla c^{j} \nabla c^{j + 1} \cdot ({| \nabla λ^{j + 1} |}^{2} \nabla^{2} λ^{j + 1} + \nabla ({| \nabla λ^{j + 1} |}^{2}) \cdot \nabla λ^{j + 1}) d x \\ + C \int_{Ω} \nabla^{2} c^{j + 1} \cdot ({| \nabla λ^{j + 1} |}^{2} \nabla^{2} λ^{j + 1} + \nabla ({| \nabla λ^{j + 1} |}^{2}) \cdot \nabla λ^{j + 1}) d x \\ \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + C \int_{Ω} {| \nabla c^{j} |}^{2} {| \nabla c^{j + 1} |}^{2} \cdot {| \nabla λ^{j + 1} |}^{2} d x \\ + C \int_{Ω} {| \nabla^{2} c^{j + 1} |}^{2} \cdot {| \nabla λ^{j + 1} |}^{2} d x \\ \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + C {‖ \nabla c^{j} ‖}_{L^{4} (Ω)}^{2} {‖ \nabla c^{j + 1} ‖}_{L^{8} (Ω)}^{2} {‖ {| \nabla λ^{j + 1} |}^{2} ‖}_{L^{4} (Ω)}^{} \\ + C {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{} {‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{2} \end{array}$

$\begin{array}{l} \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + C M_{0}^{2} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{} {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{} {‖ {| \nabla λ^{j + 1} |}^{2} ‖}_{L^{4} (Ω)}^{} \\ + M_{0} ({‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + C {‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4} \\ \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + 2 M_{0} ({‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{2}) \\ + C ({‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{4}) . \end{array}$

同理，可得

$\begin{array}{l} J_{3} \leq \int_{Ω} {| \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + \frac{1}{2} {‖ \nabla ({| \nabla c^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + M_{0} {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} \\ + C ({‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{4}) . \end{array}$

从而有

$\begin{array}{l} \frac{d}{d t} \int_{Ω} {| \nabla λ^{j + 1} |}^{4} d x + {\int_{Ω} | \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x + \frac{1}{2} {\int_{Ω} | \nabla ({| \nabla λ^{j + 1} |}^{2}) |}^{2} d x \\ \leq \frac{1}{2} {‖ \nabla ({| \nabla c^{j + 1} |}^{2}) ‖}_{L^{2} (Ω)}^{2} + 3 M_{0} ({‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}^{2}) \\ + C ({‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{4} + 1) . \end{array}$ (3.19)

我们采用与(3.19)式相同的方法，得出

$\begin{array}{l} \frac{d}{d t} \int_{Ω} {| \nabla c^{j + 1} |}^{4} d x + {\int_{Ω} | \nabla c^{j + 1} |}^{2} {| \nabla^{2} c^{j + 1} |}^{2} d x + \int_{Ω} {| \nabla ({| \nabla c^{j + 1} |}^{2}) |}^{2} d x \\ \leq 2 M_{0} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C M_{0} ({‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ λ^{j + 1} ‖}_{H^{1} (Ω)}^{2} + 1) . \end{array}$ (3.20)

$\frac{d}{d t} \int_{Ω} {| \nabla u^{j + 1} |}^{4} d x + \int_{Ω} {| \nabla u^{j + 1} |}^{2} {| \nabla^{2} u^{j + 1} |}^{2} d x + \frac{1}{2} \int_{Ω} {| \nabla ({| \nabla u^{j + 1} |}^{2}) |}^{2} d x \leq M_{0} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C ({‖ \nabla u^{j + 1} ‖}_{L^{4} (Ω)}^{4} + 1) .$ (3.21)

4) 封闭性

综合(3.5)式、(3.16)式和(3.19)式，仔细调整不等式的系数，可以找到一个适当的正常数 $α_{1}$ 使得

$\begin{array}{l} \frac{d}{d t} ({‖ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4}) + α_{1} ({‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2}) \\ + α_{1} {\int_{Ω} | \nabla λ^{j + 1} |}^{2} {| \nabla^{2} λ^{j + 1} |}^{2} d x \\ \leq \int_{Ω} {| \nabla c^{j + 1} |}^{2} {| \nabla^{2} c^{j + 1} |}^{2} d x + 11 M_{0} ({‖ Δ λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + \frac{C M_{0}}{2} {‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} \\ + C ({‖ \nabla λ^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)}^{2} + 1) . \end{array}$ (3.22)

综合(3.6)式、(3.17)式和(3.20)式，仔细调整不等式的系数，可以找到一个适当的正常数 $α_{2}$ 使得

$\begin{array}{l} \frac{d}{d t} ({‖ c^{j + 1} ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{4}) + α_{2} ({‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ c^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + α_{2} {\int_{Ω} | \nabla c^{j + 1} |}^{2} {| \nabla^{2} c^{j + 1} |}^{2} d x \\ \leq 4 M_{0} {‖ Δ c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C ({‖ \nabla c^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ c^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ λ^{j + 1} ‖}_{H^{1} (Ω)}^{2} + 1) . \end{array}$ (3.23)

综合(3.7)式、(3.18)式和(3.21)式，仔细调整不等式的系数，可以找到一个适当的正常数 $α_{3}$ 使得

$\begin{array}{l} \frac{d}{d t} ({‖ \nabla u^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ u^{j + 1} ‖}_{H^{1} (Ω)}^{2}) + α_{3} ({‖ \nabla u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + {‖ Δ u^{j + 1} ‖}_{L^{2} (Ω)}^{2}) + α_{3} {\int_{Ω} | \nabla u^{j + 1} |}^{2} {| \nabla^{2} u^{j + 1} |}^{2} d x \\ \leq 7 M_{0} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{2} (Ω)}^{2} + C ({‖ \nabla u^{j + 1} ‖}_{L^{4} (Ω)}^{4} + {‖ u^{j + 1} ‖}_{H^{1} (Ω)}^{2} + 1) . \end{array}$ (3.24)

再结合(3.22)~(3.24)式并调整系数，然后将所得不等式在0到t上进行积分，可以推导得出

$\begin{array}{l} ({‖ S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (t) ‖}_{L^{4} (Ω)}^{4}) + \int_{0}^{t} {‖ \nabla S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} d s + \int_{0}^{t} ({‖ \nabla S^{j + 1} (t) \cdot \nabla^{2} S^{j + 1} (t) ‖}_{L^{2} (Ω)}^{2}) d s \\ \leq C \int_{0}^{t} ({‖ S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (t) ‖}_{L^{4} (Ω)}^{4}) d s + 3 M_{0} \int_{0}^{t} {‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)}^{2} d s \\ + C (t + {‖ S^{j + 1} (0) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (0) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$ (3.25)

为了处理(3.25)式右端 $\nabla^{3} c^{j + 1}$ 的 $L^{2}$ 范数，我们将方程(3.1)₂重写为

$Δ c^{j + 1} = c_{t}^{j + 1} + u^{j} \cdot \nabla c^{j + 1} + (λ^{j + 1} + n_{\infty}) f (c^{j})$ (3.26)

将 $\nabla$ 作用于方程(3.26)，可以推导得出

${‖ \nabla^{3} c^{j + 1} ‖}_{L^{2} (Ω)} \leq {‖ \partial_{t} \nabla c^{j + 1} ‖}_{L^{2} (Ω)} + M_{0} ({‖ \nabla^{2} c^{j + 1} ‖}_{L^{2} (Ω)} + {‖ \nabla^{2} λ^{j + 1} ‖}_{L^{2} (Ω)}) + C M_{0} ({‖ \nabla c^{j + 1} ‖}_{L^{2} (Ω)} + {‖ \nabla λ^{j + 1} ‖}_{L^{2} (Ω)} + 1) .$ (3.27)

因此，将(3.27)式代入(3.25)式，可得

$\begin{array}{l} ({‖ S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (t) ‖}_{L^{4} (Ω)}^{4}) + (1 - 3 M_{0}^{2}) \int_{0}^{t} {‖ \nabla S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} d s + \int_{0}^{t} ({‖ \nabla S^{j + 1} (t) \cdot \nabla^{2} S^{j + 1} (t) ‖}_{L^{2} (Ω)}^{2}) d s \\ \leq C \int_{0}^{t} ({‖ S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (t) ‖}_{L^{4} (Ω)}^{4}) d s + 3 M_{0} \int_{0}^{t} {‖ \partial_{t} \nabla c^{j + 1} ‖}_{L^{2} (Ω)}^{2} d s \\ + C (t + {‖ S^{j + 1} (0) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (0) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$ (3.28)

运用 $\partial_{t} c^{j + 1} (t)$ 在 $t = 0$ 处的连续性，当 $M_{0}$ 充分小时，由(3.28)式可得

$\begin{array}{l} ({‖ S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (t) ‖}_{L^{4} (Ω)}^{4}) + \int_{0}^{t} {‖ \nabla S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} d s + \int_{0}^{t} ({‖ \nabla S^{j + 1} (t) \cdot \nabla^{2} S^{j + 1} (t) ‖}_{L^{2} (Ω)}^{2}) d s \\ \leq C \int_{0}^{t} ({‖ S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (t) ‖}_{L^{4} (Ω)}^{4}) d s + C (T_{0} + {‖ S (0) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S (0) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$ (3.29)

再对任意的 $t \in [0, T_{0}]$ ，根据Gronwall不等式我们可以得到

$({‖ S^{j + 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S^{j + 1} (t) ‖}_{L^{4} (Ω)}^{4}) \leq C (T_{0} + {‖ S (0) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla S (0) ‖}_{L^{4} (Ω)}^{4}) (1 + C T_{0} e^{C T_{0}}) .$

由上述不等式可知，当 $ε_{1}$ 和 $T_{0}$ 充分小时，(3.3)式对 $j + 1$ 成立。

第二步： ${S^{j}}_{j \geq 0}$ 的收敛性。

为了证明序列 ${S^{j}}_{j \geq 0}$ 的收敛性，我们对方程(3.1)分别取 $j + 1$ 和j，然后对两式作差得

${\begin{cases} \partial_{t} (λ^{j + 1} - λ^{j}) - Δ (λ^{j + 1} - λ^{j}) \\ = - u^{j} \cdot \nabla (λ^{j + 1} - λ^{j}) - \nabla λ^{j} \cdot (u^{j} - u^{j - 1}) - n_{\infty} \nabla \cdot (χ (c^{j}) \nabla (c^{j + 1} - c^{j})) \\ - n_{\infty} \nabla \cdot ((χ (c^{j}) - χ (c^{j - 1})) \nabla c^{j}) - \nabla \cdot (χ (c^{j}) λ^{j} \nabla (c^{j + 1} - c^{j})) \\ - \nabla \cdot ((χ (c^{j}) λ^{j} - χ (c^{j - 1}) λ^{j - 1}) \nabla c^{j}), x \in Ω, t > 0, \\ \partial_{t} (c^{j + 1} - c^{j}) - Δ (c^{j + 1} - c^{j}) \\ = - u^{j} \cdot \nabla (c^{j + 1} - c^{j}) - \nabla c^{j} \cdot (u^{j} - u^{j - 1}) - (λ^{j + 1} - λ^{j}) f (c^{j}) \\ - (f (c^{j}) - f (c^{j - 1})) (λ^{j} + n_{\infty}), x \in Ω, t > 0, \\ \partial_{t} (u^{j + 1} - u^{j}) - Δ (u^{j + 1} - u^{j}) + \nabla ({\tilde{P}}^{j + 1} - {\tilde{P}}^{j}) \\ = - u^{j} \cdot \nabla (u^{j + 1} - u^{j}) - \nabla u^{j} \cdot (u^{j} - u^{j - 1}) - (λ^{j} - λ^{j - 1}) \nabla φ, x \in Ω, t > 0, \\ \nabla \cdot (u^{j + 1} - u^{j}) = 0, x \in Ω, t > 0, \end{cases}$

其中初边值条件满足

${\begin{cases} {(λ^{j + 1} - λ^{j}, c^{j + 1} - c^{j}, u^{j + 1} - u^{j})}_{t = 0} = (0, 0, 0), x \in Ω \\ \partial_{2} (λ^{j + 1} - λ^{j}) = \partial_{2} (c^{j + 1} - c^{j}) = 0, u^{j + 1} - u^{j} = 0, x \in Γ_{B} \cup Γ_{T}, \end{cases}$

类似地，运用与证明(3.29)式相同的方法，经过一系列细致的计算，我们可以找到一个正常数 $α_{4}$ 使得

$\begin{array}{l} \sup_{0 \leq t \leq T_{0}} ({‖ S^{j + 1} (t) - S^{j} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (S^{j + 1} (t) - S^{j} (t)) ‖}_{L^{4} (Ω)}^{4}) + α_{4} \int_{0}^{t} {‖ \nabla (S^{j + 1} (t) - S^{j} (t)) ‖}_{H^{1} (Ω)}^{2} d s \\ + α_{4} \int_{0}^{t} ({‖ \nabla (S^{j + 1} (t) - S^{j} (t)) \cdot \nabla^{2} (S^{j + 1} (t) - S^{j} (t)) ‖}_{L^{2} (Ω)}^{2}) d s \\ \leq C (1 + M_{0}) T_{0} \sup_{0 \leq t \leq T_{0}} ({‖ S^{j} (t) - S^{j - 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (S^{j} (t) - S^{j - 1} (t)) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$

只需 $T_{0}$ 适当小，则存在一个常数 $θ \in (0, 1)$ ，使得对任意的 $j \geq 1$ ，有

$\begin{array}{l} \sup_{0 \leq t \leq T_{0}} ({‖ S^{j + 1} (t) - S^{j} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (S^{j + 1} (t) - S^{j} (t)) ‖}_{L^{4} (Ω)}^{4}) \\ \leq θ \sup_{0 \leq t \leq T_{0}} ({‖ S^{j} (t) - S^{j - 1} (t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (S^{j} (t) - S^{j - 1} (t)) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$ (3.30)

因此， ${S^{j}}_{j \geq 0}$ 是Banach空间 $C ([0, T_{0}]; H^{1} (Ω) \cap W^{1, 4} (Ω))$ 中的一个Cauchy序列，且 $C ([0, T_{0}]; H^{1} (Ω) \cap W^{1, 4} (Ω))$ 中存在极限函数

$S = S^{0} + \lim_{n \to \infty} \sum_{j = 0}^{\infty} (S^{j + 1} - S^{j}),$

满足

$\sup_{0 \leq t \leq T_{0}} {‖ S (t) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq \sup_{0 \leq t \leq T_{0}} \lim_{j \to \infty} \inf {‖ S^{j} (t) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq M_{0} .$

最后，用与证明收敛性类似的方法来证明系统(1.5)~(1.6)解的唯一性。故引理3.1证毕。

4. 整体存在性

4.1. 一致先验估计

为了证明定理1.1，我们在本小节中将致力于得到系统(1.5)~(1.6)解的一致先验估计。

引理4.1 假设定理1.1中所有的条件都成立。设系统(1.5)~(1.6)的解 $S : = (λ, c, u)$ 在 $[0, T]$ 上满足

$\sup_{0 \leq t \leq T} {‖ S (t) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq ε, 0 < ε < 1$ (4.1)

那么对任意的 $x \in Ω, t \in [0, T]$ ，都有 $n (x, t) \equiv λ (x, t) + n_{\infty} \geq 0, c (x, t) \geq 0, x \in Ω, t \geq 0$ 成立，并且存在两个正常数 $ε_{1}, C_{1}$ 使得对任意的 $0 < ε \leq ε_{2}$ 和 $t \in [0, T]$ 成立

$\begin{array}{l} ({‖ (λ, c, u) (\cdot, t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ, c, u) ‖}_{L^{4} (Ω)}^{4}) \\ + \int_{0}^{t} ({‖ (\nabla λ, \nabla c, \nabla u) (\cdot, s) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ, c, u) \cdot Δ (λ, c, u) ‖}_{L^{2} (Ω)}^{2}) d s \\ \leq C_{1} ({‖ (λ_{0}, c_{0}, u_{0}) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ_{0}, c_{0}, u_{0}) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$ (4.2)

证明：由假设条件(1.4)和极大值原理可知，对任意的 $x \in Ω, t \in [0, T]$ ，有 $n (x, t) \geq 0$ 和 $0 \leq c (x, t) \leq c_{\infty}$ 成立。为了得到不等式(4.2)的证明，我们将引理4.1的证明过程分为五个步骤。

第一步：S的 $L^{2}$ 估计。

首先，在方程(1.5)₁的左右两端同时乘以 $λ$ 并在 $Ω$ 上对x进行分部积分得

$\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} = \int_{Ω} (n_{\infty} + λ) χ (c) \nabla c \cdot \nabla λ d x \\ \leq n_{\infty} M_{χ} {‖ \nabla c ‖}_{L^{2} (Ω)} {‖ \nabla λ ‖}_{L^{2} (Ω)} + M_{χ} {‖ λ ‖}_{L^{4} (Ω)} {‖ \nabla c ‖}_{L^{4} (Ω)} {‖ \nabla λ ‖}_{L^{2} (Ω)} \\ \leq n_{\infty} M_{χ} {‖ \nabla c ‖}_{L^{2} (Ω)} {‖ \nabla λ ‖}_{L^{2} (Ω)} + C_{2} M_{χ} ε {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla c ‖}_{L^{2} (Ω)}^{\frac{1}{2}} {‖ \nabla λ ‖}_{L^{2} (Ω)} \\ \leq \frac{1}{2} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + \frac{C_{2}^{2} M_{χ}^{2} (n_{\infty}^{2} + ε^{2})}{2} ({‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2}) . \end{array}$

即

$\frac{d}{d t} {‖ λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} \leq C_{2}^{2} M_{χ}^{2} (n_{\infty}^{2} + ε^{2}) ({‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2}) .$ (4.3)

这里我们运用了条件(1.4)、Hölder不等式、Young不等式、Gagliardo-Nirenberg插值不等式和Sobolev嵌入定理。同理，在方程(1.5)₂的左右两端同时乘以c并在 $Ω$ 上进行分部积分得

$\frac{d}{d t} {‖ c ‖}_{L^{2} (Ω)}^{2} + 2 {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + 2 \int_{Ω} f (c) c (n_{\infty} + λ) d x = 0.$ (4.4)

最后，在方程(1.5)₃的左右两端同时乘以 $u$ 并在 $Ω$ 上进行分部积分，并利用Sobolev嵌入定理和Poincaré不等式得

$\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ u ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} = \int_{Ω} u \cdot φ \nabla λ d x \leq {‖ u ‖}_{L^{2} (Ω)} {‖ φ ‖}_{L^{\infty} (Ω)} {‖ \nabla λ ‖}_{L^{2} (Ω)} \\ \leq {(\int_{ℝ} \int_{0}^{1} {| u |}^{2} d x_{2} d x_{1})}^{\frac{1}{2}} {‖ φ ‖}_{L^{\infty} (Ω)} {‖ \nabla λ ‖}_{L^{2} (Ω)} \\ \leq C_{3} {(\int_{ℝ} \int_{0}^{1} {| \nabla u |}^{2} d x_{2} d x_{1})}^{\frac{1}{2}} {‖ φ ‖}_{L^{\infty} (Ω)} {‖ \nabla λ ‖}_{L^{2} (Ω)} \\ \leq \frac{1}{2} {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} + \frac{{(C_{3} {‖ φ ‖}_{L^{\infty} (Ω)})}^{2}}{2} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} . \end{array}$

整理上式可得

$\frac{d}{d t} {‖ u ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} \leq {(C_{3} {‖ φ ‖}_{L^{\infty} (Ω)})}^{2} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} .$ (4.5)

选择适当大的正常数 $γ_{1}, γ_{2}$ 满足 $\frac{γ_{1}}{2} \geq {(C_{3} {‖ φ ‖}_{L^{\infty} (Ω)})}^{2}$ 和 $γ_{2} \geq γ_{1} C_{2}^{2} M_{χ}^{2} (n_{\infty}^{2} + ε^{2})$ ，通过适当的调整不等式(4.3)~(4.5)的系数，可以找到一个适当的正常数 $α_{5}$ 使得

$\frac{d}{d t} (γ_{1} {‖ λ ‖}_{L^{2} (Ω)}^{2} + γ_{2} {‖ c ‖}_{L^{2} (Ω)}^{2} + {‖ u ‖}_{L^{2} (Ω)}^{2}) + α_{5} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) \leq γ_{2} {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} .$ (4.6)

第二步： $\nabla S$ 的 $L^{2}$ 估计。

首先，推导 $\nabla λ$ 的 $L^{2}$ 估计。取方程(1.5)₁与 $- Δ λ$ 的 $L^{2}$ 内积，分部积分得

$\frac{1}{2} \frac{d}{d t} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ Δ λ ‖}_{L^{2} (Ω)}^{2} = \int_{Ω} (u \cdot \nabla λ) \cdot Δ λ d x + n_{\infty} \int_{Ω} \nabla (χ (c) \nabla c) \cdot Δ λ d x + \int_{Ω} \nabla \cdot (χ (c) λ \nabla c) \cdot Δ λ d x .$ (4.7)

接下来，需要对等式(4.7)右端的三项逐一进行估计。与(3.8)式的推导过程类似，可以得出

$\int_{Ω} (u \cdot \nabla λ) \cdot Δ λ d x \leq \frac{1}{4} {‖ Δ λ ‖}_{L^{2} (Ω)}^{2} + 8 C_{4}^{4} ε^{4} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} .$ (4.8)

$n_{\infty} \int_{Ω} \nabla (χ (c) \nabla c) \cdot Δ λ d x \leq \frac{1}{4} {‖ Δ λ ‖}_{L^{2} (Ω)}^{2} + 3 n_{\infty}^{2} M_{χ}^{2} C_{5}^{2} ε^{2} {‖ Δ c ‖}_{L^{2} (Ω)}^{2} .$ (4.9)

$\begin{array}{l} \int_{Ω} \nabla \cdot (χ (c) λ \nabla c) \cdot Δ λ d x \leq \frac{1}{4} {‖ Δ λ ‖}_{L^{2} (Ω)}^{2} + ε^{2} {‖ \nabla^{3} c ‖}_{L^{2} (Ω)}^{2} + C_{7}^{4} M_{χ}^{2} ε^{2} (1 + ε^{2}) {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} \\ + C_{7}^{4} M_{χ}^{2} ε^{4} ({‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2}) . \end{array}$ (4.10)

将(4.8)~(4.10)代入(4.7)得

$\begin{array}{l} \frac{d}{d t} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + \frac{1}{2} {‖ Δ λ ‖}_{L^{2} (Ω)}^{2} \leq 2 ε^{2} {‖ \nabla^{3} c ‖}_{L^{2} (Ω)}^{2} + 2 M_{χ}^{2} ε^{2} (3 C_{5}^{2} n_{\infty}^{2} + C_{7}^{4} (1 + ε^{2})) {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} \\ + 2 M_{χ}^{2} C_{7}^{4} ε^{4} {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + 2 ε^{4} (8 C_{4}^{4} + C_{7}^{4} M_{χ}^{2}) {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (4.11)

类似地，可以推导出 $\nabla c$ 和 $\nabla u$ 的 $L^{2}$ 估计如下

$\frac{d}{d t} {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ Δ c ‖}_{L^{2} (Ω)}^{2} \leq 5 M_{f}^{2} (C_{8}^{2} ε^{2} + n_{\infty}^{2}) {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + 4 C_{8}^{2} M_{f}^{2} ε^{2} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} .$ (4.12)

$\frac{d}{d t} {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} + {‖ Δ u ‖}_{L^{2} (Ω)}^{2} \leq 16 ε^{4} {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} + 2 {‖ φ ‖}_{L^{\infty} (Ω)}^{2} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} .$ (4.13)

结合(4.11)~(4.13)式，只需取 $ε$ 适当小，使得 $M_{χ}^{2} ε^{2} (3 C_{5}^{2} n_{\infty}^{2} + C_{7}^{4} (1 + ε^{2})) \leq \frac{1}{2}$ ，并且记

$A_{1} : = \max {8 C_{4}^{4} + C_{7}^{4} M_{χ}^{2} + 2 C_{8}^{2} M_{f}^{2} + {‖ φ ‖}_{L^{\infty} (Ω)}^{2}, 3 M_{f}^{2} (C_{8}^{2} + n_{\infty}^{2}) + M_{χ}^{2} C_{7}^{4}}$ ，利用假设 $ε < 1$ ，可以得出

$\begin{array}{l} \frac{1}{2} \frac{d}{d t} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) + \frac{1}{4} ({‖ Δ λ ‖}_{L^{2} (Ω)}^{2} + {‖ Δ c ‖}_{L^{2} (Ω)}^{2} + {‖ Δ u ‖}_{L^{2} (Ω)}^{2}) \\ \leq ε {‖ \nabla^{3} c ‖}_{L^{2} (Ω)}^{2} + A_{1} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) . \end{array}$ (4.14)

第三步： $\nabla S$ 的 $L^{4}$ 估计。

首先，先对方程(1.5)₁作用 $\nabla$ ，再将得到的方程两端同乘以 $4 {| \nabla λ |}^{2} \nabla λ$ ，然后在 $Ω$ 上进行积分得

$\begin{array}{l} \frac{d}{d t} \int_{Ω} {| \nabla λ |}^{4} d x + 4 \int_{Ω} {| \nabla λ |}^{2} {| \nabla^{2} λ |}^{2} d x + 2 \int_{Ω} {| \nabla ({| \nabla λ |}^{2}) |}^{2} d x \\ = - 4 \int_{Ω} \nabla \cdot (u \cdot \nabla λ + n_{\infty} \nabla \cdot (χ (c) \nabla c) + \nabla \cdot (χ (c) λ \nabla c)) \cdot {| \nabla λ |}^{2} \nabla λ d x \\ = 4 \int_{Ω} u \cdot \nabla λ \cdot ({| \nabla λ |}^{2} \nabla^{2} λ + \nabla ({| \nabla λ |}^{2}) \cdot \nabla λ) d x \\ + 4 \int_{Ω} n_{\infty} \nabla \cdot (χ (c) \nabla c) \cdot ({| \nabla λ |}^{2} \nabla^{2} λ + \nabla ({| \nabla λ |}^{2}) \cdot \nabla λ) d x \\ + 4 \int_{Ω} \nabla \cdot (χ (c) λ \nabla c) \cdot ({| \nabla λ |}^{2} \nabla^{2} λ + \nabla ({| \nabla λ |}^{2}) \cdot \nabla λ) d x . \end{array}$ (4.15)

接下来，对(4.15)式右端的三个积分项逐一进行估计。与(3.19)式的推导过程类似，可以得到

$\begin{array}{l} 4 \int_{Ω} u \cdot \nabla λ \cdot ({| \nabla λ |}^{2} \nabla^{2} λ + \nabla ({| \nabla λ |}^{2}) \cdot \nabla λ) d x \\ \leq \int_{Ω} {| \nabla λ |}^{2} {| \nabla^{2} λ |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ |}^{2}) ‖}_{L^{2} (Ω)}^{2} + ε {‖ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + \frac{ε}{4} (C_{9}^{2} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + C_{10}^{2} {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) . \end{array}$ (4.16)

$\begin{array}{l} 4 \int_{Ω} n_{\infty} \nabla \cdot (χ (c) \nabla c) \cdot ({| \nabla λ |}^{2} \nabla^{2} λ + \nabla ({| \nabla λ |}^{2}) \cdot \nabla λ) d x \\ \leq \int_{Ω} {| \nabla λ |}^{2} {| \nabla^{2} λ |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ |}^{2}) ‖}_{L^{2} (Ω)}^{2} + ε ({‖ \nabla^{3} c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2}) \\ + \frac{n_{\infty}^{2} M_{χ}^{2} ε}{4} (C_{11}^{2} + C_{12}^{2}) {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + \frac{C_{13}^{2} ε}{4} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (4.17)

$\begin{array}{l} 4 \int_{Ω} \nabla \cdot (χ (c) λ \nabla c) \cdot ({| \nabla λ |}^{2} \nabla^{2} λ + \nabla ({| \nabla λ |}^{2}) \cdot \nabla λ) d x \\ \leq \int_{Ω} {| \nabla λ |}^{2} {| \nabla^{2} λ |}^{2} d x + \frac{1}{2} {‖ \nabla ({| \nabla λ |}^{2}) ‖}_{L^{2} (Ω)}^{2} + \frac{M_{χ}^{2} C_{14}^{2} ε}{4} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2}) \\ + \frac{1}{2} {‖ \nabla ({| \nabla c |}^{2}) ‖}_{L^{2} (Ω)}^{2} + ε ({‖ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{3} c ‖}_{L^{2} (Ω)}^{2}) . \end{array}$ (4.18)

将(4.16)~(4.18)不等式，代入(4.15)式得

$\begin{array}{l} \frac{d}{d t} \int_{Ω} {| \nabla λ |}^{4} d x + \int_{Ω} {| \nabla λ |}^{2} {| \nabla^{2} λ |}^{2} d x + \frac{1}{2} {\int_{Ω} | \nabla ({| \nabla λ |}^{2}) |}^{2} d x \\ \leq \frac{1}{2} {‖ \nabla ({| \nabla c |}^{2}) ‖}_{L^{2} (Ω)}^{2} + \frac{ε}{4} (M_{χ}^{2} C_{14}^{2} + C_{9}^{2} + C_{13}^{2}) {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + \frac{M_{χ}^{2} C_{14}^{2} ε}{4} {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + \frac{ε C_{10}^{2}}{4} {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} \\ + 3 ε {‖ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + 2 ε {‖ \nabla^{3} c ‖}_{L^{2} (Ω)}^{2} + [\frac{n_{\infty}^{2} M_{χ}^{2} ε}{4} (C_{11}^{2} + C_{12}^{2}) + ε] {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (4.19)

类似地，可以得到

$\begin{array}{l} \frac{d}{d t} \int_{Ω} {| \nabla c |}^{4} d x + \int_{Ω} {| \nabla c |}^{2} {| \nabla^{2} c |}^{2} d x + \int_{Ω} {| \nabla ({| \nabla c |}^{2}) |}^{2} d x \\ \leq 3 ε {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + \frac{ε}{4} (C_{15} + M_{f}^{2} C_{16} + M_{f}^{2} C_{17}) {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + \frac{M_{f}^{2} C_{16} ε}{4} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + \frac{C_{15} ε}{4} {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (4.20)

$\begin{array}{l} \frac{d}{d t} \int_{Ω} {| \nabla u |}^{4} d x + \int_{Ω} {| \nabla u |}^{2} {| \nabla^{2} u |}^{2} d x + \frac{1}{2} \int_{Ω} {| \nabla ({| \nabla u |}^{2}) |}^{2} d x \\ \leq 2 ε {‖ \nabla^{2} u ‖}_{L^{2} (Ω)}^{2} + \frac{ε}{4} (C_{18} + C_{19}) {‖ \nabla u ‖}_{L^{2} (Ω)}^{2} + \frac{C_{19} ε}{4} {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} . \end{array}$ (4.21)

因此，结合(4.19)~(4.21)式，取 $ε$ 适当小，使得

$\frac{n_{\infty}^{2} M_{χ}^{2} ε}{4} (C_{11}^{2} + C_{12}^{2}) + 4 ε \leq \frac{1}{4}$ ，

并且选择一个适当大的正常数 $A_{2}$ ，记

$A_{2} : = 4 \max {C_{9}^{2} + C_{13}^{2} + C_{15}, C_{10}^{2} + C_{18} + M_{χ}^{2} C_{14}^{2} + M_{f}^{2} C_{16}, C_{19} + M_{f}^{2} C_{17}}$

利用假设 $ε < 1$ ，可以得出

$\begin{array}{l} \frac{d}{d t} ({‖ \nabla λ ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla u ‖}_{L^{4} (Ω)}^{4}) + 2 {‖ \nabla λ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + 2 {‖ \nabla c \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + 2 {‖ \nabla u \nabla^{2} u ‖}_{L^{2} (Ω)}^{2} \\ \leq 2 ε {‖ \nabla^{3} c ‖}_{L^{2} (Ω)}^{2} + \frac{1}{4} ({‖ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} u ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2}) + A_{2} ({‖ \nabla u ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2}) . \end{array}$ (4.22)

第四步：一致先验估计。

首先，将方程(1.5)₂改写为

$Δ c = c_{t} + u \cdot \nabla c + λ (f (c) - f (0)) + n_{\infty} (f (c) - f (0))$ (4.23)

在方程(4.23)两端同时作用 $\nabla$ ，利用Hölder不等式、Young不等式和Sobolev嵌入定理计算可以得到

$\begin{matrix} {‖ \nabla^{3} c ‖}_{L^{2} (Ω)} = {‖ \nabla Δ c ‖}_{L^{2} (Ω)} \\ \leq {‖ \partial_{t} \nabla c ‖}_{L^{2} (Ω)} + {‖ \nabla u ‖}_{L^{4} (Ω)} {‖ \nabla c ‖}_{L^{4} (Ω)} + {‖ u ‖}_{L^{\infty} (Ω)} {‖ \nabla^{2} c ‖}_{L^{2} (Ω)} \\ + M_{f} {‖ \nabla c ‖}_{L^{4} (Ω)} {‖ λ ‖}_{L^{4} (Ω)} + n_{\infty} M_{f} {‖ \nabla c ‖}_{L^{2} (Ω)} + M_{f} {‖ c ‖}_{L^{4} (Ω)} {‖ \nabla λ ‖}_{L^{4} (Ω)} \\ \leq {‖ \partial_{t} \nabla c ‖}_{L^{2} (Ω)} + {‖ \nabla u ‖}_{L^{4} (Ω)}^{2} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + {‖ u ‖}_{H^{2} (Ω)}^{2} \\ + C_{20} M_{f} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2}) + M_{f} ({‖ \nabla λ ‖}_{L^{4} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{4} (Ω)}^{2}) \\ \leq {‖ \partial_{t} \nabla c ‖}_{L^{2} (Ω)} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} u ‖}_{L^{2} (Ω)}^{2} + M_{f} ε^{2} \\ + C_{20} M_{f} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) . \end{matrix}$ (4.24)

对方程(1.5)₂两端先作用 $\partial_{t}$ ，再与 $\partial_{t} c$ 作 $L^{2}$ 内积，得到

$\frac{d}{d t} {‖ \partial_{t} c ‖}_{L^{2} (Ω)}^{2} + {‖ \partial_{t} \nabla c ‖}_{L^{2} (Ω)}^{2} \leq M_{f} ε ({‖ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} u ‖}_{L^{2} (Ω)}^{2})$ (4.25)

取 $ε$ 适当小，使得 $C_{20} M_{f} ε \leq \frac{1}{32}$ ，将不等式(4.14)、(4.24)和(4.25)相加得

$\begin{array}{l} \frac{d}{d t} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) + \frac{3}{8} ({‖ Δ λ ‖}_{L^{2} (Ω)}^{2} + {‖ Δ c ‖}_{L^{2} (Ω)}^{2} + {‖ Δ u ‖}_{L^{2} (Ω)}^{2}) \\ \leq 3 A_{1} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) . \end{array}$ (4.26)

将不等式(4.22)、(4.24)和(4.25)相加得

$\begin{array}{l} \frac{d}{d t} ({‖ \nabla λ ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla u ‖}_{L^{4} (Ω)}^{4}) + 2 {‖ \nabla λ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + 2 {‖ \nabla c \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + 2 {‖ \nabla u \nabla^{2} u ‖}_{L^{2} (Ω)}^{2} \\ \leq \frac{1}{3} ({‖ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} u ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla^{2} c ‖}_{L^{2} (Ω)}^{2}) + 2 A_{2} ({‖ \nabla u ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2}) . \end{array}$ (4.27)

其次，在不等式(4.26)两端同乘 $\frac{α_{5}}{288 A_{1}}$ ，在不等式(4.27)两端同乘 $\frac{α_{5}}{396 A_{2}}$ ，将所得不等式与(4.6)式相加，并取 $γ_{2} = \frac{α_{5}}{2304 A_{1} A_{2}}$ ，则存在 $ε_{2} \in (0, 1)$ ，对 $\forall 0 < ε \leq ε_{2}$ 有

$\begin{array}{l} \frac{d}{d t} (γ_{1} {‖ λ ‖}_{L^{2} (Ω)}^{2} + γ_{2} {‖ c ‖}_{L^{2} (Ω)}^{2} + {‖ u ‖}_{L^{2} (Ω)}^{2}) + \frac{α_{5}}{96} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) \\ + \frac{α_{5}}{288 A_{1} A_{2}} \frac{d}{d t} ({‖ \nabla λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u ‖}_{L^{2} (Ω)}^{2}) \\ + \frac{α_{5}}{38016 A_{1} A_{2}} ({‖ Δ λ ‖}_{L^{2} (Ω)}^{2} + {‖ Δ c ‖}_{L^{2} (Ω)}^{2} + {‖ Δ u ‖}_{L^{2} (Ω)}^{2}) \\ + \frac{α_{5}}{396 A_{1} A_{2}} \frac{d}{d t} ({‖ \nabla λ ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla c ‖}_{L^{4} (Ω)}^{4} + {‖ \nabla u ‖}_{L^{4} (Ω)}^{4}) \\ + \frac{α_{5}}{198 A_{1} A_{2}} ({‖ \nabla λ \nabla^{2} λ ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla c \nabla^{2} c ‖}_{L^{2} (Ω)}^{2} + {‖ \nabla u \nabla^{2} u ‖}_{L^{2} (Ω)}^{2}) \leq 0. \end{array}$ (4.28)

最后，将不等式(4.28)从0到t积分，可知对 $\forall t \in [0, T]$ 有

$\begin{array}{l} ({‖ (λ, c, u) (\cdot, t) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ, c, u) ‖}_{L^{4} (Ω)}^{4}) \\ + \int_{0}^{t} ({‖ (\nabla λ, \nabla c, \nabla u) (\cdot, s) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ, c, u) \cdot Δ (λ, c, u) ‖}_{L^{2} (Ω)}^{2}) d s \\ \leq C_{21} ({‖ (λ_{0}, c_{0}, u_{0}) ‖}_{H^{1} (Ω)}^{2} + {‖ \nabla (λ_{0}, c_{0}, u_{0}) ‖}_{L^{4} (Ω)}^{4}) . \end{array}$ (4.29)

其中 $C_{21} > 0$ ，故引理4.1得证。

4.2. 整体存在性

根据引理4.1我们可以知道在空间 $C ([0, T]; H^{1} (Ω) \cap W^{1, 4} (Ω))$ 上，对于一些 $T > 0$ ，方程组(1.5)~(1.6)存在局部唯一解 $(λ, c, u)$ 。因此还需证明T能够拓展到 $+ \infty$ 。

首先，取 $ε_{0} : = \min {ε_{1}, ε_{2}}$ ，其中 $ε_{1}$ 和 $ε_{2}$ 都是正数，分别由引理3.1和引理4.1确定的。

然后，选取适当小的初值 $S (0)$ 满足

${‖ S (0) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq \frac{ε_{0}}{2 \sqrt{1 + C_{1}}} \leq \frac{ε_{0}}{2} \leq ε_{2},$

其中 $C_{1} > 0$ 由引理4.1给出。记方程组(1.5)-(1.6)解的最大存在时间为

$T^{*} : = \sup {t : \sup_{0 \leq s \leq t} {‖ S (s) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq ε_{0}} .$

再根据引理3.1的局部存在性结论和连续性可得 $T^{*} > 0$ 。若 $T^{*}$ 为有限时间，那么就有

$\sup_{0 \leq s \leq T^{*}} {‖ S (s) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} = ε_{0},$

这与引理4.1得到一致先验估计

$\sup_{0 \leq s \leq T^{*}} {‖ S (s) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq \sqrt{C_{1}} {‖ S (0) ‖}_{H^{1} (Ω) \cap W^{1, 4} (Ω)} \leq \frac{\sqrt{C_{1}} ε_{0}}{2 \sqrt{1 + C_{1}}} \leq \frac{ε_{0}}{2},$

相矛盾，所以 $T^{*}$ 是无限的时间。因此，引理4.1中的时间T能够拓展到 $+ \infty$ 。最后，对于不等式(1.7)的证明，可以由(4.29)式得到，从而完成了定理1.1的证明。

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